Absolute Value and Inequalities - Maths: an open, verified university mathematics reference

Why this matters¶

A bolt is specified as 20 mm long, with a tolerance of 0.05 mm. Which lengths are acceptable? A length $x$ (in mm) is acceptable when its distance from 20 is at most 0.05. We write that distance as $\abs{x - 20}$ , so the condition is $\abs{x - 20} \le 0.05$ , and this is the same as $19.95 \le x \le 20.05$ .

Calculus is full of statements of this kind: “ $x$ is close to $a$ ”, “ $f(x)$ is within a tolerance of $L$ ”. The absolute value turns “close” into an inequality, and the rules of this page turn that inequality into an interval that you can draw on the number line. The precise definition of a limit, in the Limits chapter, is written entirely in this language.

Absolute value as distance¶

We use the interval notation of Real Numbers and Intervals throughout.

In words. $\abs{x}$ is the distance from $x$ to 0 on the number line, so it is never negative. More generally, $\abs{x - a}$ is how far apart $x$ and $a$ are, whichever of the two is larger.

Example. $\abs{-3} = -(-3) = 3$ , and the distance between -1 and 4 is $\abs{-1 - 4} = \abs{-5} = 5$ .

Non-example. $\abs{x}$ does not mean “delete the minus sign”. For $x = -2$ we get $\abs{-x} = \abs{2} = 2$ , which is $-x$ , not $x$ : the formula $\abs{-x} = x$ fails for every negative $x$ .

Remark 1 (Basic properties)

For all real numbers $x$ and $y$ :

$\abs{x} \ge 0$ , and $\abs{x} = 0$ only when $x = 0$ ;
$\abs{-x} = \abs{x}$ ;
$\abs{x}$ is the larger of $x$ and $-x$ , that is, $\abs{x} = \max\{x, -x\}$ ; in particular $-\abs{x} \le x \le \abs{x}$ ;
$\abs{xy} = \abs{x}\,\abs{y}$ , and $\abs{x / y} = \abs{x} / \abs{y}$ when $y \ne 0$ ;
$\abs{x}^2 = x^2$ and $\sqrt{x^2} = \abs{x}$ ;
if $x \ge 0$ and $y \ge 0$ , then $x < y$ if and only if $x^2 < y^2$ , and $x = y$ if and only if $x^2 = y^2$ .

Properties 1–4 and the first half of 5 follow from the definition by checking the cases $x \ge 0$ and $x < 0$ (for 4, the four combinations of signs of $x$ and $y$ ). For 3: if $x \ge 0$ then $\abs{x} = x \ge -x$ , and if $x < 0$ then $\abs{x} = -x > x$ . In both cases $\abs{x}$ is the larger of the two numbers.

Properties 5 and 6 rest on the square-root remark in Real Numbers and Intervals. It states that every $y \ge 0$ has exactly one square root $\sqrt{y} \ge 0$ , and that for $s, t \ge 0$ we have $s < t \iff s^2 < t^2$ and $s = t \iff s^2 = t^2$ . For 5: $\abs{x} \ge 0$ and $\abs{x}^2 = x^2$ , so $\abs{x}$ is the one non-negative square root of $x^2$ . Property 6 is the second statement of that remark, with $s = x$ and $t = y$ .

Working with inequalities¶

Real Numbers and Intervals states trichotomy and the meaning of $\le$ . We use these order properties of the real numbers. For all real numbers $u$ , $v$ , $w$ , $s$ , $t$ and $c$ :

trichotomy: exactly one of $u < v$ , $u = v$ and $u > v$ holds; so “ $u \le v$ ” fails exactly when $u > v$ , and “ $u < v$ ” fails exactly when $u \ge v$ ;
transitivity: $u < v$ and $v < w$ imply $u < w$ ; the same holds when one of the two inequalities is $\le$ (for example $u \le v < w$ implies $u < w$ ), and $u \le v \le w$ implies $u \le w$ ;
adding the same number to both sides keeps an inequality: $u < v$ implies $u + c < v + c$ ;
adding two inequalities: $u \le v$ and $s \le t$ imply $u + s \le v + t$ , and the sum is strict, $u + s < v + t$ , if at least one of the two is strict;
multiplying both sides by a positive number keeps an inequality, and multiplying by a negative number reverses it: if $u < v$ and $c > 0$ then $cu < cv$ , but if $c < 0$ then $cu > cv$ ;
multiplying by a number that may be zero: if $u \le v$ and $c \ge 0$ then $cu \le cv$ . This holds in particular when $u < v$ , but the result is then only $\le$ : for $c = 0$ both sides are 0;
signs: a product or a quotient of two non-zero numbers is positive when they have the same sign, and negative when their signs differ.

The rules for adding and multiplying also hold when the inequality being transformed is $\le$ : the conclusion is then $\le$ (or $\ge$ when $c < 0$ ). For example, $u \le v$ and $c < 0$ give $cu \ge cv$ .

To solve a quadratic or rational inequality, we factor it and make a sign table. A factor $x - r$ is negative for $x < r$ and positive for $x > r$ , so it changes sign only at $r$ . The zeros of the factors split the number line into intervals, and on each interval every factor, and hence the whole expression, keeps one sign.

Main results¶

The first result is the translation between distances and intervals that the rest of calculus uses.

Proposition 1 (Distance inequalities as intervals)

Let $a$ and $\delta$ be real numbers. For every real number $x$ ,

\begin{aligned} &\text{(a)} && \abs{x - a} < \delta && \iff\quad a - \delta < x < a + \delta,\\ &\text{(b)} && \abs{x - a} \le \delta && \iff\quad a - \delta \le x \le a + \delta,\\ &\text{(c)} && \abs{x - a} > \delta && \iff\quad x < a - \delta \ \text{ or } \ x > a + \delta,\\ &\text{(d)} && \abs{x - a} \ge \delta && \iff\quad x \le a - \delta \ \text{ or } \ x \ge a + \delta,\\ &\text{(e)} && 0 < \abs{x - a} < \delta && \iff\quad a - \delta < x < a \ \text{ or } \ a < x < a + \delta. \end{aligned}

(2)

In particular, when $\delta > 0$ , (a) says that $x$ lies in the open interval $(a - \delta, a + \delta)$ , (b) that $x$ lies in the closed interval $[a - \delta, a + \delta]$ , and (e) that $x$ lies in the set

(a - \delta, a) \cup (a, a + \delta),

(3)

the open interval $(a - \delta, a + \delta)$ with its centre $a$ removed. This set is called the punctured neighbourhood of $a$ of radius $\delta$ .

Proof

We first treat a number $y$ in place of $x - a$ , using that $\abs{y}$ is the larger of $y$ and $-y$ , and then shift by $a$ .

By property 3 of Remark 1, $\abs{y} = \max\{y, -y\}$ . The larger of two numbers is less than $\delta$ if and only if both of them are, so

\abs{y} < \delta \iff y < \delta \text{ and } -y < \delta \iff y < \delta \text{ and } y > -\delta \iff -\delta < y < \delta .

(4)

The middle step multiplies $-y < \delta$ by -1, which reverses the inequality, and multiplying $y > -\delta$ by -1 takes us back. The same argument, with the $\le$ forms of the same rules, gives $\abs{y} \le \delta \iff -\delta \le y \le \delta$ .

Now put $y = x - a$ . Adding the same number $a$ to all three parts of $-\delta < x - a < \delta$ gives $a - \delta < x < a + \delta$ , and adding $-a$ undoes this, so the two are equivalent. This proves (a), and (b) follows in the same way with $\le$ .

Part (c) is the negation of (b): by trichotomy, $\abs{x - a} > \delta$ says exactly that $\abs{x - a} \le \delta$ fails, and “ $a - \delta \le x \le a + \delta$ ” fails exactly when $x < a - \delta$ or $x > a + \delta$ . In the same way, (d) is the negation of (a).

For (e), the condition $0 < \abs{x - a}$ says that $x \ne a$ . By property 1 of Remark 1, $\abs{x - a}$ is never negative and is 0 only when $x - a = 0$ ; conversely, if $x = a$ then $\abs{x - a} = \abs{0} = 0$ by Definition 1. So by (a), $0 < \abs{x - a} < \delta$ holds if and only if $a - \delta < x < a + \delta$ and $x \ne a$ . By trichotomy, $x \ne a$ means $x < a$ or $x > a$ , which splits the double inequality into $a - \delta < x < a$ or $a < x < a + \delta$ .

Figure 1 shows parts (a) and (e).

Two number lines, each with the points a minus delta, a and a plus delta marked. On the top line, a thick segment runs from a minus delta to a plus delta, with hollow dots at both ends. A point x inside the segment is marked, and a bracket above the line from a to x is labelled with the distance, the absolute value of x minus a. On the bottom line, the same segment has a third hollow dot at its centre a. — Figure 1:Top: the $x$ with $\abs{x - a} < \delta$ form the open interval $(a - \delta, a + \delta)$ ; the bracket marks the distance $\abs{x - a}$ from $a$ to one such $x$ . Bottom: the $x$ with $0 < \abs{x - a} < \delta$ form the punctured neighbourhood $(a - \delta, a) \cup (a, a + \delta)$ ; the hollow dot at $a$ shows that $a$ is left out.

The case $a = 0$ is used so often that it is worth stating on its own: $\abs{y} < \delta$ if and only if $-\delta < y < \delta$ , and $\abs{y} \le \delta$ if and only if $-\delta \le y \le \delta$ .

The second result says that the size of a sum is at most the sum of the sizes. Part (b) is the form for distances: going from $x$ to $y$ by way of $z$ is never shorter than going directly.

Proof

For (a) we bound $x + y$ and $-(x + y)$ separately, then use that $\abs{x + y}$ is the larger of the two.

(a) By property 3 of Remark 1, $x \le \abs{x}$ and $-x \le \abs{x}$ , and likewise $y \le \abs{y}$ and $-y \le \abs{y}$ . Adding two inequalities (the $\le$ form),

x + y \le \abs{x} + \abs{y} \qquad\text{and}\qquad -(x + y) = (-x) + (-y) \le \abs{x} + \abs{y}.

(6)

Since $\abs{x + y} = \max\{x + y, -(x + y)\}$ , again by property 3 of Remark 1, and the larger of two numbers is at most $\abs{x} + \abs{y}$ when both of them are, it follows that $\abs{x + y} \le \abs{x} + \abs{y}$ .

(b) Write $x - y = (x - z) + (z - y)$ and apply (a) to the numbers $x - z$ and $z - y$ .

\abs{x} = \abs{(x - y) + y} \le \abs{x - y} + \abs{y}, \qquad\text{so}\qquad \abs{x} - \abs{y} \le \abs{x - y}.

(7)

Exchanging the roles of $x$ and $y$ gives $\abs{y} - \abs{x} \le \abs{y - x}$ , and $\abs{y - x} = \abs{x - y}$ by property 2 of Remark 1. So both $\abs{x} - \abs{y}$ and its negative are at most $\abs{x - y}$ , and so is the larger of the two. That larger number is $\abs{\abs{x} - \abs{y}}$ by property 3 of Remark 1, which proves (c).

Worked examples¶

Example 2 (A quadratic inequality: solve

x^2 - x - 6 \le 0

)

Factor: $x^2 - x - 6 = (x + 2)(x - 3)$ . The zeros are -2 and 3.
Make the sign table.
$x < -2$ $-2 < x < 3$ $x > 3$
$x + 2$ $-$ $+$ $+$
$x - 3$ $-$ $-$ $+$
$(x + 2)(x - 3)$ $+$ $-$ $+$
The product is negative on $(-2, 3)$ and zero at -2 and 3. The inequality allows equality, so both zeros belong to the solution set.

\boxed{[-2, 3]}

(9)

Check. $x = 0$ : $-6 \le 0$ . ✓ $x = 4$ : $16 - 4 - 6 = 6$ , which is positive, so 4 is rightly excluded. ✓ $x = -2$ : $4 + 2 - 6 = 0 \le 0$ . ✓

Example 3 (A strict absolute-value inequality: solve

\abs{2x - 5} < 3

)

By Proposition 1 (a) with $a = 0$ , applied to $y = 2x - 5$ : $-3 < 2x - 5 < 3$ .
Add 5 to all three parts: $2 < 2x < 8$ .
Divide by 2, which is positive: $1 < x < 4$ .

\boxed{(1, 4)}

(10)

The distance reading gives the same set: $\abs{2x - 5} = 2\abs{x - \tfrac{5}{2}}$ , so the inequality says that $x$ is within $\tfrac{3}{2}$ of $\tfrac{5}{2}$ . The interval $(1, 4)$ has midpoint $\tfrac{5}{2}$ and half-width $\tfrac{3}{2}$ .

Check. $x = 2$ : $\abs{-1} = 1 < 3$ . ✓ The endpoint $x = 1$ gives $\abs{-3} = 3$ , which is not less than 3. ✓ $x = 5$ : $\abs{5} = 5$ , outside. ✓

Example 6 (Bounding with the triangle inequality)

Show that if $\abs{x - 2} < 1$ , then $\abs{x + 2} < 5$ and $\abs{x^2 - 4} \le 5\abs{x - 2}$ .

Write $x + 2$ in terms of the distance to 2: $x + 2 = (x - 2) + 4$ .
By Theorem 1 (a), and adding 4 to both sides of $\abs{x - 2} < 1$ ,
$\abs{x + 2} \le \abs{x - 2} + \abs{4} = \abs{x - 2} + 4 < 1 + 4 = 5 .$
(13)
A $\le$ followed by a $<$ gives $<$ (transitivity), so $\abs{x + 2} < 5$ .
Factor and use property 4 of Remark 1: $\abs{x^2 - 4} = \abs{(x - 2)(x + 2)} = \abs{x - 2}\,\abs{x + 2}$ .
Multiply $\abs{x + 2} < 5$ by $\abs{x - 2}$ . This number is $\ge 0$ (property 1 of Remark 1), but it may be 0: it is 0 at $x = 2$ . So multiplying keeps the inequality only in the form $\le$ :
$\abs{x^2 - 4} = \abs{x - 2}\,\abs{x + 2} \le 5\abs{x - 2} ,$
(14)
with equality at $x = 2$ , where both sides are 0.

\boxed{\abs{x + 2} < 5 \quad\text{and}\quad \abs{x^2 - 4} \le 5\abs{x - 2}}

(15)

Check. $x = 2.9$ : $\abs{4.9} = 4.9 < 5$ ✓, and $\abs{2.9^2 - 4} = 4.41 \le 5 \cdot 0.9 = 4.5$ . ✓

Common mistakes¶

Rigorous track¶

When is the triangle inequality an equality?

We show that $\abs{x + y} = \abs{x} + \abs{y}$ if and only if $xy \ge 0$ .

Both sides are non-negative, so by property 6 of Remark 1 they are equal if and only if their squares are. By property 5 and property 4 of Remark 1,

\abs{x + y}^2 = x^2 + 2xy + y^2, \qquad \bigl(\abs{x} + \abs{y}\bigr)^2 = x^2 + 2\abs{xy} + y^2 .

(16)

These are equal if and only if $xy = \abs{xy}$ , that is, if and only if $xy \ge 0$ .

No condition on

\delta

is needed

The proof of Proposition 1 never used the sign of $\delta$ , so the proposition holds for every real $\delta$ . For $\delta < 0$ both sides of (a), (b) and (e) are false for every $x$ . For $\delta = 0$ , (a) and (e) have no solutions, and (b) says that $\abs{x - a} \le 0$ if and only if $x = a$ . Later pages use the proposition with $\delta > 0$ , where both sides describe a genuine interval or punctured neighbourhood.

Summary¶

$\abs{x} = x$ for $x \ge 0$ and $\abs{x} = -x$ for $x < 0$ ; $\abs{x - a}$ is the distance between $x$ and $a$ .
$\abs{x - a} < \delta$ if and only if $a - \delta < x < a + \delta$ , and likewise with $\le$ (Proposition 1).
“Outside” conditions, $\abs{x - a} > \delta$ , become an or: $x < a - \delta$ or $x > a + \delta$ .
$0 < \abs{x - a} < \delta$ describes the punctured neighbourhood $(a - \delta, a) \cup (a, a + \delta)$ : the interval $(a - \delta, a + \delta)$ without its centre $a$ (Proposition 1 (e)).
Triangle inequality: $\abs{x + y} \le \abs{x} + \abs{y}$ , and $\abs{x - y} \le \abs{x - z} + \abs{z - y}$ (Theorem 1).
To solve a quadratic or rational inequality, factor, make a sign table, and check the endpoints; never divide by a quantity whose sign you do not know.

Exercises¶

Solution to Exercise 8 #

The inequality is $x^3 - 4x \ge 0$ , and $x^3 - 4x = x(x^2 - 4) = (x + 2)\,x\,(x - 2)$ , with zeros -2, 0 and 2. The sign table:

	$x < -2$	$-2 < x < 0$	$0 < x < 2$	$x > 2$
$x + 2$	$-$	$+$	$+$	$+$
$x$	$-$	$-$	$+$	$+$
$x - 2$	$-$	$-$	$-$	$+$
product	$-$	$+$	$-$	$+$

The product is positive on $(-2, 0)$ and $(2, \infty)$ , and zero at the three zeros, which the inequality allows. The solution set is $[-2, 0] \cup [2, \infty)$ . Dividing by $x$ at the start would have lost the interval $[-2, 0]$ .

Where this leads¶

These topics build on this one:

The Limit of a Function

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